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Designing my first tube preamp
Designing my first tube preamp
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Old 6th December 2019, 01:56 AM   #21
Verbstank is offline Verbstank
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Quote:
Originally Posted by rayma View Post
Usually a 12AX7 will be configured have about 1mA plate current.
The 12AU7 is often used with 5mA-10mA

Here's a more advanced follower circuit, similar to what you need. They also mention the resistor grid bias circuit.
The Valve Wizard -Cathode Follower
.

Yeah I edited my last post to say I was dumb and used the max current ratings rather than operating condition currents.


Attached an updated schematic if you are still hanging in there Values are substitutes until I work out the math on what they should be.



I had read through that Valve Wizard page previously but it makes much more sense now that you have explained a few things! So his section on Choosing the Load is exactly what I would use to find the value for the load resistor?



And because the cathode follower provides no gain, is that why you were questioning the tone circuit? Because I would not be gaining it back up before the output? If that is the case, would it be better to put the tone circuit in between the two 12ax7s and allow the 2nd 12ax7 stage to go directly to the cathode follower?
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Old 6th December 2019, 01:59 AM   #22
rayma is offline rayma  United States
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Quote:
Originally Posted by Verbstank View Post
Yeah I edited my last post to say I was dumb and used the max current ratings rather than operating condition currents.


Attached an updated schematic if you are still hanging in there Values are substitutes until I work out the math on what they should be.



I had read through that Valve Wizard page previously but it makes much more sense now that you have explained a few things! So his section on Choosing the Load is exactly what I would use to find the value for the load resistor?



And because the cathode follower provides no gain, is that why you were questioning the tone circuit? Because I would not be gaining it back up before the output? If that is the case, would it be better to put the tone circuit in between the two 12ax7s and allow the 2nd 12ax7 stage to go directly to the cathode follower?
Right, their circuit is about the same as yours, so just use their resistors.
Moving the tone won't change the total gain, but will raise the noise level.
This circuit should be pretty quiet if there's enough gain.

If you think the gain is enough as-is, go ahead with this circuit.
You may have to readjust the power supply resistor to get close to 300VDC,
but 2.5k should be a good start.

Last edited by rayma; 6th December 2019 at 02:02 AM.
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Old 6th December 2019, 02:07 AM   #23
Verbstank is offline Verbstank
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Quote:
Originally Posted by rayma View Post
Right, their circuit is about the same as yours, so just use their resistors.
Moving the tone won't change the total gain, but will raise the noise level.

If you think the gain is enough as-is, go ahead with this circuit. Should be good.
You may have to readjust the power supply resistor to get close to 300VDC,
but 2.5k should be a good start.

Man thank you so much, I feel like this was a breakthrough for understanding how to value all my resistors! So after all of this, what have I done? Dropped the output impedance (how much would I expect)? Does this affect the preamp's ability to work with tube power sections?


I think I will nix the idea of a 2 channel pre and just make it all 1 channel. If so then I could set another 12ax7 to gain post tone controls, and then feed that plate directly into that CFollower1 pdf?
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Old 6th December 2019, 02:19 AM   #24
rayma is offline rayma  United States
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Quote:
Originally Posted by Verbstank View Post
Man thank you so much, I feel like this was a breakthrough for understanding how to value all my resistors! So after all of this, what have I done? Dropped the output impedance (how much would I expect)? Does this affect the preamp's ability to work with tube power sections?

I think I will nix the idea of a 2 channel pre and just make it all 1 channel. If so then I could set another 12ax7 to gain post tone controls, and then feed that plate directly into that CFollower1 pdf?
The 12AU7 follower output impedance will be very roughly 350R, which should drive an amp ok.
You can add the 12AX7 circuit, between the volume control, and the capacitor to the grid of the 12AU7.
Then with only one channel, you would increase the power supply resistor up to around 4.7k,
but adjust the value to get 300V. Use at least a 2W resistor there.

Last edited by rayma; 6th December 2019 at 02:38 AM.
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Old 6th December 2019, 02:42 AM   #25
Verbstank is offline Verbstank
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Quote:
Originally Posted by rayma View Post
The 12AU7 follower output impedance will be very roughly 350R, which should drive an amp ok.
You can add the 12AX7 circuit, between the volume control, and the capacitor to the grid of the 12AU7.
Then with only one channel, you would increase the power supply resistor up to around 4.7k,
but adjust the value to get 300V. Use at least a 1W resistor there.

Assume I end up using all 6 triodes on either one channel or two, why would the resistor value change? Wouldn't I be using the same total plate current either way?
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Old 6th December 2019, 02:48 AM   #26
rayma is offline rayma  United States
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Originally Posted by Verbstank View Post
Assume I end up using all 6 triodes on either one channel or two, why would the resistor value change?
Wouldn't I be using the same total plate current either way?
Right now one channel has three 12AX7 halves and one 12AU7 half.
That's roughly 11mA, so 50V/.011 = 4.5k to drop 50V.
If you have two channels, current is twice as much, so halve the 4.7k to 2.4k.

If you only make one channel, you can use one entire tube for the first two stages, then a 12AX7 half for the third,
and half of the 12AU7 for the fourth. The unused sections of the last two stages wouldn't use current, since
you just ground the P/K/G terminals of the unused tube sections.

If you do make two channels, the tubes can be in a line with ch A wiring on one side, and ch B on the other side.
Use half of each tube for each channel and the wiring will work out well that way.

Last edited by rayma; 6th December 2019 at 02:55 AM.
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Old 6th December 2019, 03:58 AM   #27
PRR is offline PRR  United States
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Originally Posted by Verbstank View Post
...I was thinking that 350v would be pushing it for the 12ax7 plates....
The raw 350V will be full of ripple. You want to take 10%-30% drop in several R-C filters to clean it up. In the area of 280V. 12AX7 can take straight 300V all day long.

The tube does not take the *whole* supply voltage. It has to work in a "fair fight" with a plate resistor. To over-simplify, half voltage on resistor, half voltage on tube.

Because the 12AX7 is a high impedance we often have to let it take more voltage. In your 1.3k+100k bias, maybe 65%, so 180V. In the VERY popular Fender 1.5k+100k bias, about 70%, so 200V. Fender actually ran preamp +B over 350V on some models (cuz he had 450V available on the big bottles).

Quote:
Originally Posted by Verbstank View Post
...12ax7 tubes, taking maybe 4ma of current per plate...
The only way to approach 4mA on 12AX7 is to jam it hard-ON. But for audio we have to bias it half-on, so we can swing both ways like the strings.

Anyway we can get a close estimate of current since we know the resistor and the approximate voltage. Say 280V of B+ and 180V on tube. So it must be 100V across the resistor. Which is 100k. What is the current?
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Old 6th December 2019, 04:50 AM   #28
Verbstank is offline Verbstank
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Originally Posted by PRR View Post
The raw 350V will be full of ripple. You want to take 10%-30% drop in several R-C filters to clean it up. In the area of 280V. 12AX7 can take straight 300V all day long.

The tube does not take the *whole* supply voltage. It has to work in a "fair fight" with a plate resistor. To over-simplify, half voltage on resistor, half voltage on tube.

Because the 12AX7 is a high impedance we often have to let it take more voltage. In your 1.3k+100k bias, maybe 65%, so 180V. In the VERY popular Fender 1.5k+100k bias, about 70%, so 200V. Fender actually ran preamp +B over 350V on some models (cuz he had 450V available on the big bottles).



The only way to approach 4mA on 12AX7 is to jam it hard-ON. But for audio we have to bias it half-on, so we can swing both ways like the strings.

Anyway we can get a close estimate of current since we know the resistor and the approximate voltage. Say 280V of B+ and 180V on tube. So it must be 100V across the resistor. Which is 100k. What is the current?

1mA? 1.8mA? 2.8mA?



I don't yet grasp how Ohm's law works... Is V in the equation the amount lost across the resistor? the initial B+ amount? The 180v seen by the tube? If I take it to mean the voltage lost across the resistor that doesn't make sense because that means current is related to voltage change not voltage present...? So then I would say we are measuring the current seen by the tube at 180v? Sorry for being so beginner about this, reading the definition just doesn't seem to clear it all up for me, it's just simple algebra yet still unclear to me.


Is that then the purpose of the bias, to again adjust the voltage after the RC filtering to a better range for the tube? Again something that I've read a lot about but not 100% sure on the actual concept, I had thought of it more as a way to equalize between the anode and cathode so that there would be no signal flow if there was no input/grid change.
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Old 6th December 2019, 04:55 AM   #29
PRR is offline PRR  United States
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>> 100V across the resistor. Which is 100k. What is the current?
> 1mA? 1.8mA? 2.8mA?


100 Volts across 100k.

100k is 100,000 Ohms. (Sometimes the mix-up is not the math but the units.)

I = V/R
Click the image to open in full size.

100V/100,000 is 0.001 Amps.

0.001 Amps is 1 mA.

Since we have a lot of k-resistors, and often work in the mA range, it is useful to note that the k and m cancel out:

100V/100k = 1mA
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Last edited by PRR; 6th December 2019 at 04:59 AM.
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Old 6th December 2019, 05:04 AM   #30
Verbstank is offline Verbstank
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Quote:
Originally Posted by PRR View Post
>> 100V across the resistor. Which is 100k. What is the current?
> 1mA? 1.8mA? 2.8mA?


100 Volts across 100k.

100k is 100,000 Ohms. (Sometimes the mix-up is not the math but the units.)

I = V/R
Click the image to open in full size.

100V/100,000 is 0.001 Amps.

0.001 Amps is 1 mA.

Since we have a lot of k-resistors, and often work in the mA range, it is useful to note that the k and m cancel out:

100V/100k = 1mA

So what happens when you want to measure current at a given point? Or is that just not how that works? To measure the current at the plate, why are we measuring the current across the resistor? Because that will inherently be what we see at the plate?


Does this make sense? Sorry to turn it into an electricity lesson
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